Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is :

$A$ uniform wire of length $L$,diameter $D$,and density $\rho$ is stretched under a tension $T$. The correct relation between its fundamental frequency $f$,the length $L$,and the diameter $D$ is:

$A$ sonometer wire has a length of $114 \ cm$, between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $cm$) whose fundamental frequencies are in the ratio $1:3:4$?

$A$ sonometer wire '$A$' of diameter '$d$' under tension '$T$' having density '$\rho_1$' vibrates with fundamental frequency '$n$'. If we use another wire '$B$' which vibrates with the same frequency under tension '$2T$' and diameter '$2d$',then the density '$\rho_2$' of wire '$B$' will be:

Two strings $A$ and $B$ of lengths $L_A = 80 \text{ cm}$ and $L_B = x \text{ cm}$ respectively are used separately in a sonometer. The ratio of their densities $(d_A / d_B)$ is $0.81$. The diameter of $B$ is one-half that of $A$. If the strings have the same tension and fundamental frequency, the value of $x$ is:

The fundamental frequency of a sonometer wire is $n$. If its radius is doubled and its tension becomes half,while the material of the wire remains the same,the new fundamental frequency will be